Consider the special orthogonal group $\mathrm{SO}(m)$ equipped with the standard left-invariant Riemannian metric
$$g(X,Y)=\operatorname{tr}(X^T Y),$$
where we identify $\mathrm{T}_e\mathrm{SO}(m)$ with $\mathrm{Skew}(m)=\{X\in\mathbb{R}^{m\times m}:X+X^T=0\}$. I wish to show that a $C^2$-curve $\gamma:(-\varepsilon,\varepsilon)\to\mathrm{SO}(m)$ is a geodesic if and only if
$$\gamma''(t)^T\gamma(t)=\gamma(t)^T\gamma''(t)$$
for all $t\in(-\varepsilon,\varepsilon)$. My attempt at a solution is as follows:
As $\mathrm{SO}(m)$ is a submanifold of $\mathbb{R}^{m\times m}$, we have that the Levi-Civita connection on $\mathrm{SO}(m)$ is the tangential part of the Levi-Civita connection on $\mathbb{R}^{m\times m}$, which is simply the directional derivative through the standard identification. This implies that
$$\nabla_{\gamma'}\gamma'(t)=(\gamma''(t))^{\mathrm{tan}}\in\gamma(t)\mathrm{Skew}(m).$$
To determine the tangential part of $\gamma''(t)$ we find an expression for the orthogonal projection onto $\gamma(t)\mathrm{Skew}(m)$. By this answer we know that this is given by the mapping $X\mapsto\frac{1}{2}(X-\gamma(t)X^T\gamma(t))$, and so
$$(\gamma''(t))^{\mathrm{tan}}=\frac{1}{2}(\gamma''(t)-\gamma(t)\gamma''(t)^T\gamma(t)).$$
Rearranging we have that
$$2\gamma(t)^T\nabla_{\gamma'}\gamma'(t)=\gamma(t)^T\gamma''(t)-\gamma''(t)^T\gamma(t).$$
From this it immediately follows that $\nabla_{\gamma'}\gamma'=0$ if and only if $\gamma''^T\gamma=\gamma^T\gamma''$. As $\gamma$ is a geodesic if and only if $\nabla_{\gamma'}\gamma'=0$, the claim follows.
My question really is whether what I've done is correct or not, and whether there is some alternative way to approach this. I believe it is correct, however I am still not comfortable with the constant identifications of spaces which seems to appear all the time in differential geometry, and as such still doubt my self. Any comments are welcome.
